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100.0 mL of HCl required 25.00 mL of 0.2000 M Mg(OH)₂ to reach pH 7. What is the concentration of the HCI ? 2 HCI + Mg(OH)₂ MgCl₂ + 2 H₂O The neutralization equation is There is enough information to calculate the moles of Mg(OH)2 concentration = L Mg(OH)₂ (- a. acid i. 0.2000 Then use stoichiometry to convert moles of Mg(OH)₂ into moles of HCI mol HCI mol Mg(OH)₂ (-- p. 0.01000 b. base c. HCI d. NaOH j. 1.000 k. 0.005000 q. 0.001000 1 L mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI L HCI r. 58.31 -) = e. NaCl mol Mg(OH)₂ I. 100.0 --) = f. H₂O m. 0.1000 s. 5.831 = g. 25.00 mol/L n. 0.05000 h. 0.02500 mol Mg(OH)₂ o. 2.000 mol HCI

100.0 mL of HCl required 25.00 mL of 0.2000 M Mg(OH)₂ to reach pH 7. What is the concentration of the HCI ? 2 HCI + Mg(OH)₂ MgCl₂ + 2 H₂O The neutralization equation is There is enough information to calculate the moles of Mg(OH)2 concentration = L Mg(OH)₂ (- a. acid i. 0.2000 Then use stoichiometry to convert moles of Mg(OH)₂ into moles of HCI mol HCI mol Mg(OH)₂ (-- p. 0.01000 b. base c. HCI d. NaOH j. 1.000 k. 0.005000 q. 0.001000 1 L mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI L HCI r. 58.31 -) = e. NaCl mol Mg(OH)₂ I. 100.0 --) = f. H₂O m. 0.1000 s. 5.831 = g. 25.00 mol/L n. 0.05000 h. 0.02500 mol Mg(OH)₂ o. 2.000 mol HCI

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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100.0 mL of HCI required 25.00 mL of 0.2000 M Mg(OH)2 to reach pH 7. What is the concentration of the HCI ? 2 HCI + Mg(OH)₂ MgCl₂ + 2 H₂O The neutralization equation is There is enough information to calculate the moles of Mg(OH)2 ( L Mg(OH)₂ (- concentration = Then use stoichiometry to convert moles of Mg(OH)₂ into moles of HCI mol HCI a. acid i. 0.2000 p. 0.01000 mol Mg(OH)₂ (- b. base c. HCI d. NaOH mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI j. 1.000 k. 0.005000 q. 0.001000 1 L mol HCI LHCI r. 58.31 = mol Mg(OH)2 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 = g. 25.00 mol/L n. 0.05000 h. 0.02500 mol Mg(OH)2 o. 2.000 mol HCI
Transcribed Image Text:100.0 mL of HCI required 25.00 mL of 0.2000 M Mg(OH)2 to reach pH 7. What is the concentration of the HCI ? 2 HCI + Mg(OH)₂ MgCl₂ + 2 H₂O The neutralization equation is There is enough information to calculate the moles of Mg(OH)2 ( L Mg(OH)₂ (- concentration = Then use stoichiometry to convert moles of Mg(OH)₂ into moles of HCI mol HCI a. acid i. 0.2000 p. 0.01000 mol Mg(OH)₂ (- b. base c. HCI d. NaOH mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI j. 1.000 k. 0.005000 q. 0.001000 1 L mol HCI LHCI r. 58.31 = mol Mg(OH)2 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 = g. 25.00 mol/L n. 0.05000 h. 0.02500 mol Mg(OH)2 o. 2.000 mol HCI
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